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三校联考高三数学试题参考答案第1页共7页“德化一中、永安一中、漳平一中”三校协作2022—2023学年第一学期联考高三数学试题参考答案一、选择题：本题共8小题，每小题5分，共40分．在每小题给出的四个选项中，只有一项是符合题目要求的．1．D2．B3．C4．D

5．C6．B7．A8．D二、选择题：本题共4小题，每小题5分，共20分．在每小题给出的选项中，有多项符合题目要求。全部选对的得5分，有选错的得0分，部分选对的得2分．9．ABC10．AD11．AD12．BD三、填空题：本题共4小题，每小

题5分，共20分．13．21yx14．615．2226,0{}ee16．435四、解答题：本题共6小题，共70分．解答应写出文字说明，证明过程或演算步骤．17.解：（Ⅰ）∵向量���→=（2a﹣c，b）与向量���→=（cosC，cos

B）共线，∴（2a﹣c）cosB﹣bcosC＝0，········································································2分即(2sinsin)cossincosACBBC，

∴2sinAcosB＝sin（B+C）＝sinA，∵sinA≠0，∴cosB=������，···············································································4分∵B∈（0，π），∴B=������；·

·············································································5分（Ⅱ）由已知1333sin244ABCSacBac，所以3ac························

········6分由余弦定理得222cos33acac，所以226ac······································7分解得3ac，················································

········································9分又因为B=������，所以ABC为正三角形．·························································10分三校联考高三数学试题参考答案第2页

共7页18.解：（Ⅰ）当���=���时，������=���������−���，所以������=���，··········································1分因为������=���������−���①，所以当�

��≥���时，������−���=���������−���−���−���②，······················2分①-②得������=���������−���������−���−���

，所以������=���������−���+���，········································3分所以������+���������−���+���=���������−���+���+���������−���+���

=���������−���+���������−���+���=���，·····························································5分所以������+���是首项为2，公比为2的等比数

列，所以������+���=���⋅������−���，所以������=������−���；·······················································6分（Ⅱ）由（Ⅰ）知，����

��=���，������=���，所以������=������=���，������=������=���，···················7分设������的公差为���，则������=������+���−���⋅���，所以���=���，··············

······················8分所以������=������+���−���⋅���=���，·····································································9分所以

������������=���������−���=���⋅������−���，设数列���⋅������的前���项和为������，所以������=���×���+���×������+���×����

��+⋯+���⋅������③，���������=���×������+���×������+���×������+⋯+���⋅������+���④，③-④得−������=���+������+������+⋯+������−���⋅������+���=������−�

��������−���−���⋅������+���=���−���⋅������+���−���，所以������=���−���⋅������+���+���，··············································

························10分又因为数列���的前���项和等于���+���+���+⋯+���=������+������，································11

分所以������������的前���项和为nT=���−���⋅������+���−������+������+���.····································12分19.（Ⅰ）证明：∵AD

//BC，BC=12AD，Q为AD的中点，∴四边形BCDQ为平行四边形，∴CD//BQ······················································2分∵∠ADC=90°，∴∠AQB=90°，即QB⊥AD，又∵平面PAD⊥平面ABCD，平

面PAD∩平面ABCD=AD，∴BQ⊥平面PAD，∵BQ平面MQB，∴平面MQB⊥平面PAD；···············································5分三校联考高三数学试题参考答案第3页共7页（Ⅱ）解：∵PA=PD，

Q为AD的中点，∴PQ⊥AD，∵平面PAD⊥平面ABCD，且平面PAD∩平面ABCD=AD，∴PQ⊥平面ABCD，···6分如图，以Q为原点建立空间直角坐标系，则(0,0,0)Q，(1,0,0)A，(0,0,3)P，(0,3,0)B，(1,3,0)C，设P

MPC，且01，得(,3,33)M，所以(,3,33)QM，(0,3,0)QB，···········································7分设

平面MBQ法向量为(,,)mxyz，由00mQMmQB，得3(33)030xyzy，令3x，则1(3,0,)m，·········

·························································8分由题意知平面BQC的一个法向量为(0,0,1)n，··············································9分

∵二面角M-BQ-C为60°，∴21||||1cos60213()nmnm，解得12，·····································

························································11分∴133(,,)222QM，∴||QM72.······················

····································12分20.解：（Ⅰ）由题意可知π3COE，则扇形COE的面积为211π232π623S，························

·············3分//OCAB，则AFOFOC，且2tanAF，············································4分∴梯形OCBF的面积为2122432432tantan

S，··································6分∴1222π436tanSSS，3tan3OAABOAB，∴π6ABO，三校联考高三数学试题参考答案第4页共7页又π3EOF，所以π

π63，∴3tan,33；·····················································································8分（Ⅱ）设26tany，3tan,33

，222223sin13sin126sin26sinsinsiny，且13s2in,2，············9分记为锐角，且3sin3，当π6时，0y，此时函数26tany单调递减，当π3时，

0y，此时函数26tany单调递增，································11分∴当3sin3时，6cos3，y取最小值，S取最大值，此时sin2tancos2.···································

············································································12分21.解：（Ⅰ）连接EF1，EF2，由题意知12,32FEFc

，······························1分设12,tan,sin,cosbbcEFFcaa，121223,23,2cbEFEFEFEFaaa，即226232,3,baabaa又解得6,3ab

，椭圆C的方程为22163xy··········································································4分（Ⅱ）（i）当切线与坐标轴垂直时,交点坐标为(2,2),90,A

OBOAOB，····························································································

·····················5分(ii)当切线与坐标轴不垂直时，设切线为11(0),(,),ykxmkAxy22(,),Bxy三校联考高三数学试题参考答案第5页共7页由圆心到直线距离为2222,22,1mdmkk············

·····························6分联立切线方程与椭圆方程消去y整理得222(21)4260,kxkmxm∴2121222426,2121kmmxxxxkk，······································

························7分2222121212122366(1)()021mkxxyykxxkmxxmk，OAOB，···························

··································································9分综上所述，OAOB；（Ⅲ）当切线与坐标轴直时，4OAOB，当切线与坐标轴不垂直时,由（Ⅱ）知,2,OAOBOAOBAB····

············10分2221222(1)(82)121kkABkxxk，22,223144tkABtt令则，······················································11分当且仅当时22k等

号成立，32OAOB，综上所述，OAOB的最大值为32.···························································12分22.解：（Ⅰ）由()fx定义域为0,x，且2111xaxaxx

aafxxaxxx，·······································1分令0fx得，1x或xa，①当01a时，0,xa，()0fx，fx单调递增，,1x

a，()0fx，fx单调递减，1,x，()0fx，fx单调递增，···········································2分三校联考高三数学试题参考答案第6页共7

页②当1a时，()0fx，fx在0,单调递增，···········································3分③当1a时，0,1x，()0fx，fx单调递增，1,xa，()0fx，fx单调递减

，,xa，()0fx，fx单调递增，·············································4分综上，当01a时，fx在区间0,a，1,上单调递增，fx在区间,1a上单调递减；

当1a时，fx在区间0,上单调递增；当1a时，fx在区间0,1，,a上单调递增，fx在区间1,a单调递减；（Ⅱ）由已知，214ln2gxxaxx，则22444axa

xxxagxxxxx，函数gx有两个极值点1212,xxxx，即240xxa在0,上有两个不等实根，···········································

······································································5分令24hxxxa，只需00240haha，故04a，·············

·······················6分又124xx，12xxa，···········································································

·····7分所以2212111222114ln4ln22gxgxxaxxxaxx2212121214lnln2xxaxxxxln8aaa，·········

······8分要证1210lngxgxa，即证ln810lnaaaa，只需证1ln20aaa，令1ln2maaaa，0,4a，·················································

···········9分则11ln1lnamaaaaa，因为2110maaa恒成立，所以ma在0,4a上单调递减，又110m，12ln20

2m，由零点存在性定理得，01,2a使得00ma，即001lnaa，所以00,aa时，0ma，ma单调递增，0,4aa时，0ma，ma单调递减，三校联考高三数学试题参考答案第7页共7页则0000

max1ln2mamaaaa0000011123aaaaa，···············11分∵0013yaa在01,2a上显然单调递增，∴00111323022aa

，∴0ma，即1210lngxgxa，得证．···············································12分获得更多资源请扫码加入享学资源网微信公众号www.xiangxue100.com